# In what sense is the kernel of a group homomorphism a universal arrow in a comma category?

Let $K$ denote the kernel of the group homomorphism $f : G \rightarrow H$. Let $i : K \rightarrow G$ denote the inclusion homomorphism.

Then evidently $(K, i)$ is a universal arrow in the comma category $(S \downarrow G)$ for some functor $S$.

Question: How so? What is this $S$?

Attempt: I know that if $T$ is another group and $j$ another mapping that goes from $T$ to $G$ with the property that $j(T) \in K$, then there exists a unique mapping $j' : T \rightarrow K$ s.t. $i \circ j' = j$. Namely, $j'$ is just $j$ with codomain narrowed to $K$.

• I know of how it's a universal arrow for $(\Delta \downarrow [0,f:G\rightrightarrows H])$, but I'm unfamiliar with one you're looking for. Can you be more specific about the context in which you ran across this claim? – Malice Vidrine Mar 18 '17 at 9:06
• @MaliceVidrine: The context is from Categories for the Working Mathematician, when the author states "the kernel of a homomorphism is a universal, more exactly, a universal for a suitable contravariant functor." A few pages before, the author talks about universals in the context of comma categories, and so I'm trying to parse his claim specifically within the context of a comma category. – user1770201 Mar 18 '17 at 11:12
• Hm, the claim that it's a contravariant functor makes it all the more puzzling. And yet, it's Mac Lane, so I can't really doubt it... – Malice Vidrine Mar 19 '17 at 2:07
• @MaliceVidrine: Is what you're referring to by $(\Delta \downarrow [0,f:G\rightrightarrows H])$ the characterization presented by Andrej Bauer? – user1770201 Mar 19 '17 at 8:50
• Not quite, unless I'm misreading his answer. In the characterization I know, the comma category is over the data $\mathbf{Group}\overset{\Delta}{\longrightarrow}\mathbf{Group}^{\bullet\rightrightarrows\bullet}\overset{(f,0)}{\longleftarrow}\mathbf{1}$, where $\Delta$ takes every group to the constant directed multigraph on groups. This is just bluntly using the equaliser adjunction in the special case where one of the morphisms is the zero morphism. – Malice Vidrine Mar 19 '17 at 9:27

The inclusion of a kernel is simply a universal element of the contravariant functor $F:=\mathrm{Hom}_{f,0}(-,G)$ sending a group $L$ to the set of homomorphisms $L\to G$ vanishing upon composition with $f$. It looks to me like this is what MacLane had in mind. This can be viewed as an initial object of the comma category $\mathbf{Gp}^{\mathrm{op}}/F$ (EDIT:) that is, the category of natural transformations from representables into $F$. This is isomorphic, using Yoneda, to the category $*\downarrow F$ of elements of $F$, and since $F$ is a subfunctor of the representable on $G$, we see that $*\downarrow F$ is the corresponding subcategory of the slice category $*\downarrow yG=\mathbf{Gp}\downarrow G$, which are two other descriptions of the appropriate category given on this page.
• Is $\mathbf{Gp}^{op} / F$ just a different notation for $(\mathbf{Gp}^{op} \downarrow F)$? And is there a special name for contravariant functors of form $\text{Hom}_{f,0}(-,G)$? – user1770201 Mar 20 '17 at 7:19
• Or is $\mathbf{Gp}^{op} / F$ precisely the subcategory of $(\mathbf{Grp} \downarrow G)$ that Andrej Bauer was speaking of? (Here I'm taking $\mathbf{Gp}$ to be synonymous with $\mathbf{Grp}$). – user1770201 Mar 20 '17 at 7:45
• It seems to me more like the kernel is a universal element of the coma category $(\mathbf{*} \downarrow F )$ where $F$ is the functor defined by Kevin and $\mathbf{*}$ is the set with one element. How do you manage to introduce the category $\mathbf{Gp}^{op}$ in the first operand of the comma? – almaus Feb 11 '18 at 20:42
• @almaus That's yet a third isomorph of the categories Andrej and I suggested. What I didn't make clear in my original answer is that this comma is with respect to the Yoneda embedding $\mathbf{Gp}^{\mathrm{op}}\to [\mathbf{Gp},\mathbf{Set}]$, so an object is a natural transformation $yK\to F$, that is, an element of $F(K)$. – Kevin Arlin Feb 12 '18 at 1:26
The arrow $i$ is universal among all arrows $j$ into $G$ which equalize $f$ and $0 : G \to H$. So it is universal in the subcategory of the slice over $G$ on arrows that equalize $f$ and $0$. The answer seems to be that $i$ is universal in a full subcategory of the slice over $G$ (which is a comma category of course), but not in a straight comma category, unless there's some trick to present a full subcategory of a slice as a comma.
• So you're saying that the setting is some subcategory of $(\mathbf{Grp} \downarrow G)$ but you aren't sure how to characterize this subcategory as its own comma (or coslice) category? – user1770201 Mar 19 '17 at 8:46